Question: Is ${953439}$ divisible by $3$ ?
Explanation: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {953439}= &&{9}\cdot100000+ \\&&{5}\cdot10000+ \\&&{3}\cdot1000+ \\&&{4}\cdot100+ \\&&{3}\cdot10+ \\&&{9}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {953439}= &&{9}(99999+1)+ \\&&{5}(9999+1)+ \\&&{3}(999+1)+ \\&&{4}(99+1)+ \\&&{3}(9+1)+ \\&&{9} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {953439}= &&\gray{9\cdot99999}+ \\&&\gray{5\cdot9999}+ \\&&\gray{3\cdot999}+ \\&&\gray{4\cdot99}+ \\&&\gray{3\cdot9}+ \\&& {9}+{5}+{3}+{4}+{3}+{9} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${953439}$ is divisible by $3$ if ${ 9}+{5}+{3}+{4}+{3}+{9}$ is divisible by $3$ Add the digits of ${953439}$ $ {9}+{5}+{3}+{4}+{3}+{9} = {33} $ If ${33}$ is divisible by $3$ , then ${953439}$ must also be divisible by $3$ ${33}$ is divisible by $3$, therefore ${953439}$ must also be divisible by $3$.